JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Inverse Trigonometric Functions

All

x

satisfying the inequality

(cot^{- 1} x)^{2} - 7 (cot^{- 1} x) + 10 > 0

, lie in the interval:

Answer: (B)

Solution

Put

y = cot^{- 1} x

. The inequality

y^{2} - 7 y + 10 > 0

factors as

(y - 2) (y - 5) > 0

, so

y < 2

y > 5

. Since

cot^{- 1} x

lies in

(0, π)

, only

cot^{- 1} x \in (0, 2)

survives, and as

cot^{- 1}

is decreasing this means

x > cot 2

, i.e.

x \in (cot 2, \infty)

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